In the figure there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles.
I need to find the total area of the circles.
I know this is going to have something to do with summation as a value approaches infinity, but I'm not entirely sure how to approach the problem. Here's what I have so far:
I know that the radius of the central inscribed circle is $ \frac{\sqrt{3}}{6} $. As such, the area of the first circle is $$ A = \pi\left(\frac{\sqrt{3}}{6}\right)^2. $$ Because there are three "branches" of infinite circles, I'm assuming that the answer will look something like: $$ A = \pi\left(\frac{\sqrt{3}}{6}\right)^2 + 3 \sum_{1}^{\infty}\text{something}.$$
Since the length of the equilateral triangle is 1, let its median length $x = \frac{\sqrt(3)}{2}$. So, the radius of the first circle is $\frac{x}{3}$. Hence, its area is $A_1 = {\pi}(\frac{x}{3})^2$ The radius of the next three identical circles will be $\frac{x}{9}$. The radius of each progressing circle will be a third of the previous one. So, the sum of the areas of these circles is
$A = A_1+3\sum_{n=1}^{\infty}\frac{\pi x^2}{3^{2n+2}}$ Now, considering the fact that $x=\frac{\sqrt(3)}{2}$,
$3\sum_{n=1}^{\infty}\frac{\pi x^2}{3^{2n+2}} = 3\sum_{n=1}^{\infty}\frac{9 \pi}{4*3^{2n+2}} = \frac{\pi}{4}\sum_{n=1}^{\infty}\frac{1}{9^n}$. Considering the convergence of the geometric with r<1, the total area is given by
$A = A_1 + \frac{\pi}{4} \sum_{n=1}^{\infty} \frac{1}{9^n} = \frac{\pi}{12} + \frac{\pi}{32} = \frac{11 \pi}{96}$. Thus, the total area of the circles is $\frac{11 \pi}{96}$